Question

Não sei resolver, me ajudem!
Simplificando a equação √x .∛x².x² / x⁴ onde x > 0; obtém-se:

a) 1/ ⁶√x⁵
b) 1/ ⁵√x⁶
c) 1 / x

Answer 1

$\mathtt{\sqrt{x}.\sqrt[3]{{x}^{2}}.\dfrac{{x}^{2}}{{x}^{4}} = {x}^{ \frac{1}{2}}. {x}^{ \frac{2}{3}}. {x}^{2 - 4} }$

$\mathtt{{x}^{\frac{1}{2}}.{x}^{\frac{2}{3}}.{x}^{-2} = {x}^{ \frac{1}{2} + \frac{2}{3} - 2 } = {x}^{ \frac{3 + 4 - 12}{6} } } \\ = \mathtt{ {x}^{ - \frac{5}{6} } }$

$= \huge\boxed{\boxed{\mathtt{ \frac{1}{\sqrt[6]{ {x}^{5} } } }}}$

$\huge\boxed{\boxed{\mathtt{Alternativa~a}}}$

Answer 2

Resposta:

Alternativa a)

Explicação passo-a-passo:

$\displaystyle \frac{\sqrt{x} \sqrt[3]{x^2}.x^2 }{x^4} =\frac{x^{\frac{1}{2}}.x^{\frac{2}{3}}.x^2 }{x^4}= \frac{x^{(\frac{1}{2}+\frac{2}{3}+2)}}{x^4} =\frac{x^{(\frac{3+4+12}{6} )}}{x^4} =\frac{x^{(\frac{19}{6} )}}{x^4} =x^{(\frac{19}{6}-4 )}=x^{(\frac{19-24}{6} )}=x^{-\frac{5}{6}}=\frac{1}{\sqrt[6]{x^5} }$